CODEWITHSUNDEEP
haskellocamloperator-precedence
cswannabe
cswannabe

Reputation: 95

Equivalent of Haskell's $ operator in OCaml

Is there an equivalent to Haskell's $ operator in OCaml, or do I have to rely on brackets? See for example,

multiplyByFive 5 + 1 -- 26

but

multiplyByFive $ 5 + 1 -- 30

Upvotes: 8

Views: 800

Answers (3)

axr
axr

Reputation: 416

In OCaml, you can use the application operator (added in OCaml 4.01) to achieve the same.

multiplyByFive @@ 5 + 1
- : int = 30

The application operator carries right precedence so the right side of the operator is evaluated first, similar to Haskell's application operator ($). You might also want to look into the pipeline operator (|>) in OCaml which is a reverse application operator similar to the (&) operator in Haskell.

Upvotes: 7

octachron
octachron

Reputation: 18912

The standard library defines both a right-to-left application operator @@

let compose h g f x = h @@ g @@ f x

and left-to-right application operator |>:

let rev_compose f g h x = x |> f |> g |> h

with the expected associativity (right for @@ and left for |>).

Upvotes: 16

Lhooq
Lhooq

Reputation: 4441

You're looking for @@

# let multiplyby5 a = 5 * a;;
val multiplyby5 : int -> int = <fun>
# multiplyby5 5 + 1;;
- : int = 26
# multiplyby5 @@ 5 + 1;;
- : int = 30

Upvotes: 7

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